# How to prove $[w]_{\mathcal B} = \begin{bmatrix} \langle w, u_1\rangle\\ \vdots\\ \langle w,u_n \rangle\\ \end{bmatrix}$ [closed]

Let $V$ be a complex inner product space, $\mathcal B=\{u_1,...,u_n \}$ an ordered orthonormal basis for $V$, and $w\in V$. Prove $$[w]_{\mathcal B} = \begin{bmatrix} \langle w, u_1\rangle\\ \vdots\\ \langle w,u_n \rangle\\ \end{bmatrix}$$

## closed as off-topic by kingW3, André 3000, Trevor Gunn, Namaste, JackNov 22 '17 at 1:57

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Since $\mathcal B$ is a basis, given $w\in V$ we may write: $$w = \sum_{i=1}^{m}a_i u_i$$ where all $a_i\in \mathbb F$. It remains to prove that $a_i = \langle w,u_i\rangle$ for every $i = 1,\cdots, n$, because the entries of the matrix $[w]_{\mathcal B}$ is just the coordinates of $w$ with respect to $\mathcal B$. Therefore, given $i\in \{1,\cdots, n\}$, we evaluate the following:
$$\langle w,u_i\rangle = \langle \sum_{j=1}^{m}a_ju_j,u_i\rangle = \sum_{j=1}^ma_j\langle u_j,u_i\rangle = a_i\langle u_i,u_i\rangle = a_i,$$ since the basis is orthogonal and normalized. This proves our statement: $a_i = \langle w,u_i \rangle.$